Note that \(N\)  is nilpotent and simply connected by [1]} (warning: the definition of a `complex analytic group' in [1]} includes the assumption of being connected). Moreover, \(N\)  coincides with the radical of \((G,G)\)  by [1]}. In particular, \(N\)  is integral.
Note that \(\mathfrak {n}=[\mathfrak {g},\mathfrak {r}]=[\mathfrak {g},\mathfrak {g}]\cap \mathfrak {r}\) , where \(\mathfrak {r}\)  is the solvable radical of \(\mathfrak {g}\)  [4]}. It is easy to see that the solvable radical of \([\mathfrak {g},\mathfrak {g}]\)  coincides with \([\mathfrak {g},\mathfrak {g}]\cap \mathfrak {r}\)  and therefore with \(\mathfrak {n}\) . Thus, \(\mathfrak {n}\)  is the Lie algebra of \(N\) .
Since \(G\)  and \(R\)  are the integral subgroups of \(G\)  corresponding to \(\mathfrak {g}\)  and \(\mathfrak {r}\) , and \(\mathfrak {n}\)  is an ideal, it follows that \((G,R)\)  is the integral subgroup corresponding to \([\mathfrak {g},\mathfrak {r}]\)  [5]}. Then \([\mathfrak {g},\mathfrak {r}]=\mathfrak {n}\)  implies that \((G,R)=N\) .
